Question

A fusion reaction used to produce neutron beams,

${}_1^{2}D{+}_1^{2}D{\to }_2^{3}He{+}_0^{1}n$

Assuming that the kinetic energy before the fusion is negligible compared with the energy released, calculate the neutronkinetic energy after fusion.

Short answer

The kinetic energy of the neutron is$3.267MeV$

Step-by-step solution

Given data

The atomic mass of.${}_1^2\text{D},m_{\text{D}}=2.014102\text{u}$

The atomic mass of${}_2^3\text{He},m_{\text{He}}=3.016029\text{u}$

The atomic mass of${}_0^1\text{n},m_{\text{n}}=1.008665\text{u}$

Concept of Total kinetic energy

The energy released in this decay depends upon the mass difference between the parent and daughter nucleus.

We can calculate the kinetic energy Q of the emitted particle using the formula:

$Q=(m_p-m_D)c^2$

In this formula${\text{m}}_P$ is the mass of the parent nucleus. And${\text{m}}_D$ is the mass of the daughter nucleus.

Calculate the Kinetic Energy

The formula for kinetic energy Q is:

$Q=(m_p-m_D)c^2$

For the given reaction it can be written as follows:

localid="1658465570922" $Q=(2m_D-m_n-m_{He})c^2$

Substituting values in the above equation, we get,

$\begin{array}{c}Q=(2\times 2.014102\text{u}-1.008665\text{u}-3.016029\text{u})c^2\\ =3.51\times {10}^{-3}u\times 931.5MeV/u\\ =3.267MeV\end{array}$$\begin{array}{c}Q=(2\times 2.014102\text{u}-1.008665\text{u}-3.016029\text{u})c^2\\ =3.51\times {10}^{-3}u\times 931.5MeV/u\\ =3.267MeV\end{array}$

The kinetic energy of the neutron is .$3.267MeV$